Derivatives of an Inverse Function

The derivative of an inverse function provides the slope of the tangent line to the inverse function's graph. It has a specific relationship to the derivative of the original function. If a function f(x) is differentiable and has an inverse function f-1(x), and if f'(g(x)) ≠ 0 where g(x) = f-1(x), then the derivative of the inverse function is given by a specific formula.

If g(x) = f-1(x) is the inverse of a differentiable function f(x), the formula for the derivative of the inverse function is:

g'(x) = 1 ÷ f'(g(x))

This means the derivative of the inverse function at a point 'x' is the reciprocal of the derivative of the original function evaluated at the corresponding point on the original function (which is g(x), the y-value on the inverse, or the x-value on the original function).

Alternatively, you can use the relationship f(g(x)) = x and differentiate both sides using the Chain Rule: f'(g(x)) * g'(x) = 1. Solving for g'(x) gives the same formula.

The derivatives of the inverse trigonometric functions (arcsin x, arccos x, arctan x, etc.) are derived using the formula for the derivative of an inverse function (g'(x) = 1 ÷ f'(g(x))) and basic trigonometric identities.

Here are the formulas for the most common ones:

• d/dx (arcsin x) = 1 ÷ √(1 - x2) (for -1 < x < 1)
• d/dx (arccos x) = -1 ÷ √(1 - x2) (for -1 < x < 1)
• d/dx (arctan x) = 1 ÷ (1 + x2) (for all real x)

The formulas for arccsc x, arcsec x, and arccot x can also be derived similarly.

No, they are generally not the same. The formula g'(x) = 1 ÷ f'(g(x)) shows the relationship. You don't simply take the inverse of the derivative 1 ÷ f'(x); instead, you evaluate the reciprocal of the derivative at the corresponding point on the original function (g(x)).